after looking at this question I came to think on one particular case. I'm wondering if maybe I've missed something on the way. If anyone could give it a look that would be great:

We start by considering a subset $H=\lbrace\frac{p}{q}\in\mathbb Q| p\text{ is even and }q\text{ is odd}\rbrace$.

Naturally $0=\frac{0}{1}\in\mathbb Q$, and for any $\frac{p}{q}\in H$: $\frac{-p}{q}\in H$. Also, if $\frac{p_1}{q_1},\frac{p_2}{q_2}\in H$ then $p_1q_2+p_2q_1$ is even, and $q_1q_2$ is odd, so $H$ is indeed a sub-group of $\mathbb Q$ (sorry for going into too much detail, as I said- I'm trying to make sure I didn't miss anything)

So- what's $\mathbb{Q}/H$? Let's see what we've got:
Let $r_1,r_2\in\mathbb{Q}$ be arbitrary. Write them down as $r_1=2^{t_1}\frac{p_1}{q_1}$ and $r_2=2^{t_2}\frac{p_2}{q_2}$ with $p_1,p_2,q_1,q_2$ all odd integers, and $t_1,t_2\in\mathbb{Z}$

We know that
$$r_1=r_2\mod H\iff r_1-r_2\in H\iff \frac{2^{t_1}p_1q_2-2^{t_2}p_2q_1}{q_1q_2}\in H$$
WLOG assum $t_1\geq t_2$ then we have
$$2^{t_2}\frac{2^{t_1-t_2}p_1q_2-p_2q_1}{q_1q_1}\in H$$
From here on I notice two cases:

Case 1: If $t_2>0$ than this is always true- meaning that any two elements of $\mathbb{Q}$ with a positive diadic valuation are congruent under $H$ (since $t_1\geq t_2>0$). EDIT- Not very suprising, as any element of positive diadic valuation is $0_{\mathbb Q/H}$, by definition (Thanks to Brian for mentioning this).

Case 2: (and this is what's got me baffled) If $t_2\leq 0$ we need $2^{t_1-t_2}p_1q_2-p_2q_1$ to be an even integer, such that $2^{-t_2}$ divides it, for this number to be in $H$. Since $p_2q_1$ is odd, it must hold that $2^{t_1-t_2}=1$ so $t_1=t_2$. Also we have that $2^{-t_2}|p_1q_2-p_2q_1$.

Basically this is where I got stuck- what's the deal with Case 2? What kind of a group is this? Am I completely off track somewhere? or is this maybe a known result?

If anyone can point me at some direction here I would be very thankful.

What kind of answer do you want to «what's $\mathbb Q/H$?»? It can be described as $\operatorname{colim}_{q\text{ odd}}\mathbb Z_{2q}$ with the colimit taken over odd numbers $q$ ordered by divisibility (I think). Does that qualify as an answer?
–
Mariano Suárez-Alvarez♦Jul 5 '11 at 20:48

On the other hand, if $0 < k < m < 2^{n+1}$, and $k$ and $m$ are odd, $\frac{m}{2^n} - \frac{k}{2^n} = \frac{m-k}{2^n} \notin H$, since the highest power of $2$ that can divide $m-k$ is $2^n$. It follows that $A_n/H = \{[\frac{2k+1}{2^n}]:0 \le k < 2^n\}$, where the $2^n$ elements listed are all distinct. It's also clear that the elements of $A_n/H$ are precisely the elements of order $2^n$ in $G$. I'm no algebraist and have never dealt with the $2$-Prüfer group before, but if I understand the Wikipedia description correctly, Jyrki is correct, and that's what we have here.